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From the center C of hyperbola (x^2)/(a^...

From the center `C` of hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1` , perpendicular `C N` is drawn on any tangent to it at the point `P(asectheta,btantheta)` in the first quadrant. Find the value of `theta` so that the area of ` C P N` is maximum.

Text Solution

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Equation of tangent at `P(4 sec theta, 3 tan theta)` is
`(3 sec theta) x-(4 tan theta)y-12=0`
`therefore" "CN=(12)/(sqrt(9 sec^(2)theta+16 tan^(2)theta))`
Equation of normal at P is
`4x cos theta +3y cot theta=25`
`therefore" "CM=(25)/(sqrt(16cos^(2)theta+9 cot^(2) theta))`
`therefore" Area of triangle CPN"=(1)/(2)xxCMxxCN`
`=(1)/(2)xx(25xx12)/(sqrt(144+81"cosec"^(2) theta+256sin^(2)theta+144))`
`=(150)/(sqrt((9 "cosec"theta-16 sin theta)^(2)+576))`
`le(150)/(sqrt(0+576))`
`therefore" Maximum area of triangle CPN"=(150)/(24)=(25)/(4)` sq. units
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